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p. k x,B. k x,A. Solute molecule. Solvent molecule. A. B. x B. x B →0. x B →1. Microscopic view about Raoult law and Henry Law. Gas A,B y A ,y B. Liquid A, B x A ,X B. Ideal solution. The solvent and solute both follow Raoult Law. Binary system. Scuba diving. - PowerPoint PPT Presentation
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Microscopic view about Raoult law and Henry Law
xB
kx,A
Ap
kx,B
Bp
A B
AAA xkp ,x
AAA xpp
BBx,B xkp
BBB xpp
xB→0 xB →1
p
Solvent molecule
Solute molecule
Ideal solution
Binary system *A A A AP Py P x
LiquidA, BxA,XB
GasA,ByA,yB
The solvent and solute both follow Raoult Law
*B B B BP Py P x
*
*
( ln) ( ) ( , ) ln
( , ) ln
( , ) ln
( ) ln
AA A
A A
A
A
Pys gas T P RT
P
P xT P RT
P
T l RT x
T RT x
( ln) ( ) ( ) lnB B Bs g T RT x
* *A A A By x if P P
* *A A A By x if P P
Vaporizable solutes follow Henry Law
kX,B(kPa)(298K)/H20
H2 N2 O2 CO2 CH4
7.12x106 8.68x106 4.40x106 1.66x106 4.18x106
4 17.9 10 . .kPa kg mol
2
2 4 1
4 1
1
21
7.1 10 . .
2.9 10 .
0.29 .
oO
m
P kPam
k kPa kg mol
mol kg
mmol L
25 , sea level, O℃ 2 in air PO2=21kPa
m BBp k m
Scuba diving
Ideal diluted solution
*p p xp p k x A AA B x,B B
Solvent follow Raoult’s law
)/ln()(),( ppRTTpT AAA
AAA xRTppRTT ln)/ln()(
AA xRTpT ln),(*
Solute follow Henry’s law
)/ln()(),()1( ppRTTpT BBB
BxB xRTpkRTT ln)/ln()(
BB xRTpT ln),(*
P
xA
kA*AP
Chemical potential of B, X=1, follow Henry Law
ln)sln()sln( O22 RT 2x
)/( Ommln)sln()sln( O22 RT
)/( Occln)sln()sln( O22 RT
Chemical potential of reference/standard state ：
ln)()sln( OO2 RTg )/( O
, pK xh
)/( OO, pmK mh
)/( OO, pcK ch
ln)()sln( OO2 RTg
ln)()sln( OO2 RTg
Solute in diluted solution
4.5 Colligative properties of diluted solution
B
*
AA
*
A xpppp The solute is not volatile
Freezing-point depression
The decrease of vaporizing pressure
f f BT k m f
*
ff TTT
b b BT k m *
bbb TTT
solute
solvent
The elevation of boiling point
Osmosis: osmotic pressure
Freezing-point depression
(l) (s) Tf* (sln) (s) Tf
1(sln) = 1*(s)
R
TG
RT
Gx mfusmfus )/(
lnOO
1
1(l) + RT ln x1 = 1
(s)
RT ln x1 = –[1(l) – 1
(s) ]
p
mfus
p T
TG
RT
x
)/(1ln
O
1
2
O
RT
HmfusHG
ff
mfus
TTR
Hx
11ln
O
1
f
f
T
T
mfusx
T
dT
R
Hx
*
1
2
O
1 1ln
X2 is small lnx1=ln(1– x2) x2
ff
ff
ff TT
TT
TT
11 Tf should be small, too
f
f
mfus TRT
Hx
2
O
2
mKmH
MRTT f
mfus
ff
O
1
2
1
2
21
2
n
n
nn
n
2
f
f
T
T
111
12 mMMn
Mn
A
Am,fus
2*
ff
)(M
H
TRk
freezing point lowering coefficients
1K mol kg 。
Note: diluted solution, separate pure solid (solvent)
Boiling-point elevation
bb
mvap
TTR
Hx
11ln
O
1
mKmH
MRTT b
mvap
bb
O1
2
A
Am,vap
2*
bb
)(M
H
TRk
b b BT k m *
bbb TTT
(boiling point elevation coefficients ）， Unit: kgmolK 1 bk
Osmosis phenomenon and osmosis pressure
( ln)A s* ( )A l
mT
VP
Semipermeable membrane
AA((slnsln)=)=AA((ll) +) + RT RT lnln x xAA <<AA*(*(ll))
P↑, μ ↑
*( ln, , ) ( )A A As x p l
Osmosis pressure
* ( , )A l P
ln(1 )B Bx x
*
*
( ln, , )
( , ) ln
( , ) lnA
A A
A A
P
A
P m
s x P
l P RT x
l P V dP RT x
lnP
m A
P
V dP RT x
( )m BV P P RTx
Osmosis pressure: meaning, measurement and application
B
m
xRT cRT
V
Van’t Hoff equation
gh
20℃ ， mB=0.001mol.kg-
1
Δp=0.053Pa;ΔTf=0.002K; π=2400Pa
Colligative properties of diluted solution: Application examples
Mr. Tian Wen-fei Mr. Tian Wen-zhi
4.6 Real solution: activity of solute and solvent
i i i ix a x
*ii i
P
P x Or by Henry Law
ii
i i
P
k x Judged by Raoult law
γcs2
γacetone
By R.L.
Determination of activity of solvent
RT
Va m
1,1ln)4(
(1) p1 = p1* a1
ff
mfus
TTR
Ha
11ln)2(
O
1
bb
mvap
TTR
Ha
11ln)3(
O
1
Example 1: Activity of water in aqueous solution
16025 .fus mH J mol * 273.2
258.2
f
f
T K
T K
* 5 3 1, 1.8 10 .m waterV m mol
P0, Aqueous solution Tf=– 15℃
Question: a(water)=? Osmosis pressure of the solution.
Answer:
ff
mfus
TTR
Ha
11ln
O
11543.0
258
1
273
16025
R
a1=0.857
PaaV
RT
m
751
1,
1012.2108.1
1543.0298314.8ln
Example 2: Acetic acid(A)/benzene(B) solutionxA 0 0.0835 0.2973 0.6604 0.9931 1.000
pA /Pa 1535 3306 5360 7293 7333
pB /Pa 35197 33277 28158 18012 466.6
*
18012 / 315791.507
0.3396B B
BB B B
a P
x P x
18012 / 676230.7845
0.3396B A
BB B A
a P
x k x
P
xA0 1
40000Pa
PA
PB
By R.L
By H.L
*
5360 / 73330.731
0.6604A A
AA A A
a P
x P x
By R.L:
Pa
dx
dpK
B
B
xBxh
B
676230069.0
6.466
lim0
,,
mixG=RT(xA ln aA + xB ln aB)= – 1167J
Homeworks Y:P101:15; P104:17,18 P106: 22 P110:23 A: P190: Exe7.19(b); Problem7.1 P192: application7.23
Term seminar report :
May 21:
May 23:
May 30:
June 04:
June 06:10 min/person
